sgu 499. Greatest Greatest Common Divisor 因数打表

最新推荐文章于 2018-05-20 20:18:21 发布

原创最新推荐文章于 2018-05-20 20:18:21 发布 · 288 阅读

0 ·

CC 4.0 BY-SA版权

sgu !!!!!!!!!! 同时被 2 个专栏收录

50 篇文章

订阅专栏

水--------------------

24 篇文章

订阅专栏

本文介绍了一个计算问题，即如何找出一组正整数中任意两个数的最大公约数的最大值。通过输入一系列人的友谊编号，利用质因数分解和哈希表记录每个质因数出现的次数来解决此问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

499. Greatest Greatest Common Divisor

Time limit per test: 0.5 second(s)
Memory limit: 262144 kilobytes

input: standard
output: standard

Andrew has just made a breakthrough in sociology: he realized how to predict whether two persons will be good friends or not. It turns out that each person has an inner friendship number (a positive integer). And the quality of friendship between two persons is equal to the greatest common divisor of their friendship number. That means there are prime people (with a prime friendship number) who just can't find a good friend, and

Wait, this is irrelevant to this problem. You are given a list of friendship numbers for several people. Find the highest possible quality of friendship among all pairs of given people.

Input

The first line of the input file contains an integer n (

) — the number of people to process. The next n lines contain one integer each, between 1 and

(inclusive), the friendship numbers of the given people. All given friendship numbers are distinct.

Output

Output one integer — the highest possible quality of friendship. In other words, output the greatest greatest common divisor among all pairs of given friendship numbers.

Example(s)

sample input	sample output
4 9 15 25 16	5

#include <bits/stdc++.h>

using namespace std;

int v[1000005];

int ans=1;
void f(int x){
    if(v[x]>=2){
        ans=ans>x?ans:x;
    }
}

void solve(){
    int n;
    scanf("%d",&n);

    for(int i=1;i<=n;i++){
        int a;
        scanf("%d",&a);
        for(int i=1;i*i<=a;i++){
            if(i*i==a){
                v[i]++;
                f(i);
            }
            else{
                if(a%i==0){
                    v[a/i]++;
                    v[i]++;
                    f(i);
                    f(a/i);
                }
            }
        }
    }

    printf("%d",ans);
    return;
}

int main(){
    solve();
    return 0;
}